If $C$ is a category of fibrant objects, is its associated $\infty$-category idempotent complete, i.e. is it accessible? If this is not always true, besides from the case when it is an $n$-category for finite $n$, when is it true?
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2 Answers
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If the associated $\infty$-category $C$ is stable, one may consider all possible Verdier quotients, each of which would define a new structure of category of fibrant objects. The fact that all these Verdier quotients are idempotent-complete is equivalent to the vanishing of $K$-theory of $C$ in degree $-1$. Therefore, whenever you start from a singular algebraic variety of positive dimension, some Verdier quotient of the category of perfect complexes on the given variety $X$ gives an example of a category of fibrant objects which is idempotent-complete (as it consists of bounded complexes of vector bundles) but whose localization is not idempotent-complete, because $K_{-1}(X)$ is not zero.
This means that, even if the underlying category is very nice, essentially anything may happen, unless we are considering a Verdier quotient of a category whose $K$-theory vanishes in degree $-1$. This vanishing always occurs in the presence of a bounded $t$-structure. See the paper arXiv:1610.07207 by Antieau, Gepner and Heller.
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No, that is not always true. One family of examples is discussed in this answer by Lennart Meier, but it can fail even more subtly. Consider a dual example of the category of finite simplicial sets which is a category of cofibrant objects. It is idempotent complete and weak equivalences are closed under retracts, but its associated $(\infty, 1)$-category is not idempotent complete due to Wall's finiteness obstruction.
A positive result is that it works if limits of towers of fibrations exist and the limit of such tower is a fibration, acyclic if all fibrations in the tower are acyclic. In that case the associated $(\infty, 1)$-category is countably complete. Unfortunately, I don't know any criterion that does not impose infinite limits.
